User talk:Ikosarakt1
Welcome Hi, welcome to Googology Wiki! Thanks for your edit to the Expansion page. Please leave a message on my talk page if I can help with anything! -- FB100Z (Talk) 12:01, May 30, 2012 Whoa — haven't visited this site for a while. Thanks for all your contributions here. I wasn't aware that the googology community was big enough that this wiki is still active! If you need anything deleted, protected, blocked or stuff, just ask me. FB100Z • talk • 18:20, June 17, 2012 (UTC) Hiya. Please check out Forum:MathJax + GoogolBot and let me know what you think. FB100Z • talk • 05:07, November 10, 2012 (UTC) I wrote a comment there. Ikosarakt1 (talk) 10:21, November 10, 2012 (UTC) Hello, Ikosarakt1, First, my apologizes for editing your page, "Hyper-Operator", without asking you, and... "Tetration" (function that, I've been thinking about since my early 11 years, when, in mathematics, I viewn the "raise" (or "power") function, (the result of multiply a number by itself as times so the "exponent" tells...). And... ¿Why may not think about a further function that would consist of "raising" "to" itself a number, so times as the "X" coeficient tells? Well. The difference, the NAME I had gaven that supposed function when I referred to it. The "hyper-power", or the "super-power"... My question is, then, this: When "raising" a number (a) to itself (b) times (a^a^...^a), does "Tetration" solves it from the bottom to up (((a^a)^a)^a) (that, certainly, it is what my "Super-power" makes...), or in another hand, solves it from the up to botton (a^(a^(a^a)))?. In the first case, as computer programmer with an experiencie (modesty aside, jeee jieee..., well...), I found three faults in your pseudo-code, faults that I will put in striked. Well... In the fourth line ("result := b")... If b is the number of times that a operates by itself (in further re-calls to the, for instance, recursive function, hyper(a, b, c)...), the initial result (when b IS NOT bigger than 1) I think, should be a'''. Then, line 4 would be "result := '''a". Whell. In the 6th line, I'd seen that hyper(..,..,..) is re-called with an "n" value of n'. This could cause an infinite bucle, when the function WILL NOT TERMINATE with a DEFINITIVE result, until "n" coeficient has 0 value. Good. If you pass "n" to the n parameter, this parameter ALWAYS will have the same value: n. Then n would NEVER be 0. My logical think is, when hyper(..,..,..) is a RECURSIVE function, and the '''n'_"ation" operations consists in repeating 'n-1'_"ation"s operations of a over itself, my correction, in concert, is to pass 'n-1 '(instead of '''n...) in the "n" coeficient. (6th row: result := hyper(result, b, n-1)...). Finaly, the third fault, in the same 6th row I'd seen, is that b''' is passed in the "b" coeficient of hyper(a,b,n), instead of '''a. If a operates BY ITSELF in the recursive n-1_"ation" functions, I think the correct code is to passing a''' to this coeficient, because the result is operated by '''a each time you repeat the n-1_"ation" operation when you are solving your n_"ation"... Then, the 6th row: result := hyper(result, a, n-1) Raul Du Mamarrach (talk) 01:17, November 15, 2012 (UTC) function hyper(a'', ''b, n''): '''if' n'' = 0: '''return' a'' + ''b result := b''' '''repeat b'' - 1 '''times': result := hyper(result, b''', '''n) return result Thank you for this edit, it seems to be true. As for tetration, it can be solved from right to left, ((...((a^a)^a)^a)...)^a (with b terms) ... is a mere a^(a^(b-1)), but a tetrated to b = a^(a^(a^(a...(a^a)...), you can feel the difference. Ikosarakt1 (talk) 10:16, November 15, 2012 (UTC) Can you delete that page? Can you delete Class 4(Category page)? because it is actually exponent level. no content... wrong content? or... Jiawheinalt (talk) 12:17, January 14, 2013 (UTC) Not need. It is page which need to fill by content as Class 1, Class 2 and Class 3, because many Bowers' zillions exists in this range. Ikosarakt1 (talk) 12:47, January 14, 2013 (UTC) ok. Jiawheinalt (talk) 11:26, January 15, 2013 (UTC) Can you delete Trestrigintillion? Tretrigintillion has aleardy exist and i did not know it existed. And i created the page with the s beetween the s and the t. Thanks(formal). Jiawheinalt (talk) 08:02, January 19, 2013 (UTC) I've seen that spelling of this number may vary. I put a redirect on tretrigintillion because wikipedia uses this with s. Ikosarakt1 (talk) 08:15, January 19, 2013 (UTC) That template got a gap http://googology.wikia.com/wiki/Template:Illiard Jiawheinalt (talk) 12:22, February 3, 2013 (UTC) This template must have only redirects on zillion (-illion) pages, we recently estabilish to merge zillions with zilliards. Ikosarakt1 (talk) 20:32, February 3, 2013 (UTC) The template shall be restored. Jiawheinalt (talk) 11:37, February 14, 2013 (UTC) Can you convert this to math jax 10100 & 10 & 10 ~ {10,3 / 2} Jiawheinalt (talk) 11:20, February 4, 2013 (UTC) :\(\lbrace10,100\rbrace\&10\&10\approx\lbrace10,3/2\rbrace\) -- I want more 11:22, February 4, 2013 (UTC) Use \& to indicate & in MathJax. Ikosarakt1 (talk) 11:34, February 4, 2013 (UTC) thx to both. Jiawheinalt (talk) 11:39, February 4, 2013 (UTC) What is the page? Millian Jiawheinalt (talk) 12:30, February 8, 2013 (UTC) Deleted. Ikosarakt1 (talk) 13:01, February 8, 2013 (UTC) :Protip: If you want to preserve source-less pages, do not edit it. Otherwise it will get attention and get deleted. I know. It's partly a joke. Also I tried to delete that page but I was too slow. — I want more 13:06, February 8, 2013 (UTC) these are the template contains the illian series. http://googology.wikia.com/wiki/Template:IllianSeries Jiawheinalt (talk) 13:05, February 8, 2013 (UTC) List of prime numbers http://primes.utm.edu/lists/small/1000.txt Jiawheinalt (talk) 00:45, February 9, 2013 (UTC) You can see first 10000 primes if you change 1000 to 10000 in the address. By the way, I recently wrote a program that sorts out all primes up to a given number and counts that. For example, up to 100000000 there exists 5761455 primes. Ikosarakt1 (talk) 07:20, February 9, 2013 (UTC) prefixes Is the prefixes by jim blowers be considered as si prefixes? Jiawheinalt (talk) 12:10, February 17, 2013 (UTC) No. These prefixes is not accepted in the international system. Ikosarakt1 (talk) 12:49, February 17, 2013 (UTC) BOX_M vs. pentatri Hey, would you mind posting your proof comparing BOX_M to pentatri? FB100Z • talk • 23:28, February 18, 2013 (UTC) Well, I want to show that \(BOX\_\widetilde{M}\) < {3,4,2,1,2} < Pentatri = {3,3,3,3,3}. For a start, I bound \(\widetilde{R}\) above by {3,4,3,2}. I will use Saibian's LAPL on some expressions, e.g. \((n \uparrow\uparrow\uparrow 3) \uparrow\uparrow n < (n \uparrow\uparrow\uparrow 3) \uparrow\uparrow (n \uparrow\uparrow\uparrow 2) < n \uparrow\uparrow\uparrow 5\). Go: Line 1. \(n\$ < (n^n) \uparrow\uparrow (n^n) < n \uparrow\uparrow\uparrow 3\). Line 2. \(n¥ < ({}^{n!}(n\$)) \uparrow \cdots \uparrow ({}^{n!}(n\$)) \uparrow ({}^{n!}(n\$)) = {}^{n!}(n\$) \uparrow\uparrow n < ((n \uparrow\uparrow\uparrow 3) \uparrow\uparrow (n^n)) \uparrow\uparrow n < (n \uparrow\uparrow\uparrow 5) \uparrow\uparrow n < n \uparrow\uparrow\uparrow 7\). Line 3. \(n\widetilde{¥} < {}^{n\$}(n\$) \uparrow\uparrow n < ((n \uparrow\uparrow\uparrow 3) \uparrow\uparrow (n \uparrow\uparrow\uparrow 3)) \uparrow\uparrow n < (n \uparrow\uparrow\uparrow 5) \uparrow\uparrow n < n \uparrow\uparrow\uparrow 7\). Line 4. \(n£ < ((n \uparrow\uparrow\uparrow 7) \uparrow\uparrow (n \uparrow\uparrow\uparrow 7)) \uparrow\uparrow n < (n \uparrow\uparrow\uparrow 9) \uparrow\uparrow n < n \uparrow\uparrow\uparrow 11\). Line 5. There are some weirdness. \(A_1\) seems defined as function, but hasn't place by argument. I guess that it is properly written as \(A_1(n)\). So: \(A_1(n) < n \uparrow\uparrow\uparrow 11\). \(A_2(n) < n \uparrow\uparrow\uparrow 13\). \(A_3(n) < n \uparrow\uparrow\uparrow 15\). In general: \(A_k(n) < n \uparrow\uparrow\uparrow (2k+9)\). Line 6. \(M_k(n) = \{n,k+1,1,2\}\) Line 7. Same comments as for line 5. \(k_1(n) < \{n \uparrow\uparrow\uparrow (n \uparrow\uparrow\uparrow 12), n \uparrow\uparrow\uparrow 11,1,2\} < \{3,n \uparrow\uparrow\uparrow 12,1,2\}\). \(k_x(n) < \{3,n \uparrow\uparrow\uparrow (x+11),1,2\}\) Line 8. \(G£ < \{3,66,1,2\} < \{3,3,2,2\}, k_{G£} < \{3,4,2,2\}, k_{k_{G£}} < \{3,5,2,2\}, \widetilde{R} = k_{k_{k \cdots {k_{k_{k_{G£}}}}}} < \{3,G£+3,2,2\} < \{3,4,3,2\}\). Line 9. \(G£ \uparrow^{\widetilde{R}} G£ < \{3,66,1,2\}. M_1 < \{3,3,3,\{3,67,1,2\}\} < \{3,4,1,1,2\}\). Line 10. \(M_2 < \{3,5,1,1,2\}\). \(M_3 < \{3,6,1,1,2\}\). \(M_n < \{3,n+3,1,1,2\}\). Line 11. \(BOX\_\widetilde{M} < M_{M_1+1} < \{3,\{3,4,1,1,2\}+4,1,1,2\} < \{3,4,2,1,2\}\). Ikosarakt1 (talk) 11:34, February 19, 2013 (UTC) @FB100Z, I do mind? Jiawheinalt (talk) 13:13, February 19, 2013 (UTC) Shall we convert the Saibian's Cascading-E numbers to math jax? Jiawheinalt (talk) 11:36, March 1, 2013 (UTC) Nope. I noticed Cloudy's comment about over MathJaxing our wiki, and decided don't use it on the simple expressions. Ikosarakt1 (talk) 15:56, March 1, 2013 (UTC) \(no \leftarrow{no}\), just practicing the multitask. Jiawheinalt (talk) 10:21, March 4, 2013 (UTC)